7. The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour. (1986  4 Marks)
Ans : 66.17 mm, 0.65
Solution :
TIPS/Formulae :
P_{total }= p_{A} + p_{B}
Molecular weight of CH_{3}OH = 12 + 3 + 16 + 1 = 32
Molecular weight of C_{2}H_{5}OH= 24 + 5 + 16 + 1 = 46
According to Raoult’s law
P_{total} = p_{1} + p_{2}
where P_{total} = Total vapour pressure of the solution
p_{1} = Partial vapour pressure of one component
p_{2} = Partial vapour pressure of other component
Again, p_{1} = Vapour pressure × mole fraction
Similarly, p_{2} = Vapour pressure x mole fraction
Mole fraction of CH_{3}OH =
Mole fraction of ethanol =
NOTE THIS STEP : Thus now let us first calculate the partial vapour pressures, i.e., p_{1} and p_{2} of the two component.
Partial vapour pressure of CH_{3}OH(p_{1})= 88.7 × 0.49 = 43.48 mm
Partial vapour pressure of C_{2}H_{5}OH(p_{2}) = 44.5 × 0.51 = 22.69 mm
∴ Total vapour pressure of the solution = 43.48 + 22.69 mm = 66.17 mm
Mole fraction of CH3OH in vapour =
8. The vapour pressure of a dilute aqueous solution of glucose (C_{6}H_{12}O_{6}) is 750 mm of mercury at 373 K. Calculate (i) molality and (ii) mole fraction of the solution. (1989  3 Marks)
Ans : 0.7503 mol/kg, 0.9868
Solution :
TIPS/Formulae :
x_{2} (solute) = 1 – 0.9868 = 0.0132
= 0.7503 mol kg^{–1}
9. The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A nonvolatile nonelectrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance? (1990  3 Marks)
Ans : 65.25
Solution :
TIPS/Formulae :
According to Raoult’s law,
Here, pº = 640 mm
p = 600 mm
w= 2.175 g
= 39.0
m = ?
M = 78
Substituting the various values in the above equation for Roult's law :
10. The degree of dissociation of calcium nitrate in a dilute aqueous solution, containing 7.0 g. of the salt per 100 gm of water at 100ºC is 70%. If the vapour pressure of water at 100ºC is 760 mm, calculate the vapour pressure of the solution. (1991  4 Marks)
Ans : 746.3 mm Hg
Solution :
TIPS/Formulae : First find moles of Ca(NO3)2 and water.
Then use the expression to find vapour pressure of solution.
Let initially 1 mole of Ca(NO_{3})_{2} is taken
Degree of dissociation of Ca(NO_{3})_{2} = 70/100 = 0.7
Ionisation of Ca(NO_{3})_{2} can be represented as
∴ Total number of moles in the solution at equilibrium = (1 – 0.7) + 0.7 + 2 × 0.7 = 2.4
No. of moles when the solution contains 1 gm of calcium nitrate instead of 1 mole of the salt
11. Addition of 0.643 g of a compound to 50 ml. of benzene (density : 0.879 g/ml.) lowers the freezing point from 5.51ºC to 5.03ºC. If Kf for benzene is 5.12 K kg mol–1, calculate the molecular weight of the compound. (1992  2 Marks)
Ans : 156.056
Solution :
TIPS/Formulae :
Given Wt. of benzene (solvent),
W = Volume × density = 50 × 0.879 = 43.95 g
Wt. of compound (solute), w = 0.643 g
Mol. wt. of benzene, M = 78; Mol. wt. of solute, m = ?
Depression in freezing point, DTf = 5.51 – 5.03 = 0.48
Molal freezing constant, K_{f} = 5.12
Now we know that,
12. What weight of the nonvolatile solute, urea (NH_{2} – CO – NH_{2}) needs to be dissolved in 100g of water, in order to decrease the vapour pressure of water by 25%? What will be the molality of the solution? (1993  3 Marks)
Ans : 18.52 m
Solution :
TIPS/Formulae :
Here, w and m are wt. and molecular wt. of solute, W and M are wt. and molecular weight of solvent
p = Pressure of solution; pº = Normal vapour pressure
Let the initial (normal) pressure (pº) = p
∴ Pressure of solution =
m = 60, M = 18, W = 100 gm
13. The molar volume of liquid benzene (density=0.877 g mL^{–1}) increases by a factor of 2750 as it vaporises at 20°C and that of liquid toluene (density=0.867 g mL^{–1}) increases by a factor of 7720 at 20°C. A solution of benzene and toluene at 20°C has a vapour pressure of 46.0 Torr. Find the mole fraction of benzene in the vapour above the solution. (1996  3 Marks)
Ans : 0.73
Solution :
TIPS/Formulae :
Volume of 1 mole of liq. benzene =78/0.877
Volume of 1 mole of toluene = 92/0.867
In vapour phase,
At 20°C, for 1 mole of benzene,
Similarly for 1mole of toluene,
As we know that, PV = nRT
Total vapourpressure = 46 torr = 46/760 = 0.060 atm
Thus, 0.060 = 0.098 X_{B} + 0.029 (1 – X_{B})
⇒ 0.060 = 0.098 X_{B} + 0.029 – 0.029 X_{B}
⇒ 0.031 = 0.069 X_{B}
X_{B} + X_{T} = 1
X_{T} = 1– 0.45 = 0.55 (in liquid phase)
142 docs66 tests

142 docs66 tests


Explore Courses for JEE exam
